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I want to prove the following

If $S$ is an open subset of a perfect metric space $(M,d)$ then $\text{clr}S$ is perfect.

By definitoin $S$ is perfect if and only if

$$S=\text{der}S,$$

where $\text{der}S$ is the derived set or the set of all accumulation points of $S$. This is equivalent to

$$S\subset \text{der}S\,\land \text{der}S\subset S.$$

We know that

$$\text{clr}S=S\cup\text{der}S,\tag{1}$$

which implies that $\text{der}S\subset S$ if and only if $\text{clr} S = S$. This means that $S$ is perfect if and only if

$$S\subset\text{der} S\,\land\,\text{clr}S=S.$$

Since we want to show that $\text{clr} S$ is perfect then we should show that

$$\text{clr}S \subset\text{der}\,\text{clr}S \,\land\, \text{clr}\,\text{clr}S=\text{clr}S$$

The $\text{clr}\,\text{clr}S=\text{clr}S$ is trivial as $\text{clr}S$ is a closed set in $M$. It remains to show that

$$\text{clr}S \subset\text{der}\,\text{clr}S.$$

Choose any $x\in S$ and let $r>0$ be given. We show that $T:=B_M(x,r)\cap S$ contains a $y\ne x$. Since $S$ is open then $T$ is open and there is a ball $B_M(x,\epsilon)\subset T$. Since $T\subset S\subset M$ then $x\in M$ and $M$ is perfect so there is a $y\ne x$ in $M$ such that $y\in B_M(x,\epsilon)\cap M = B_M(x,\epsilon)$, which means that there is a $y\ne x$ such that $y\in T$. This implies that $x\in \text{der} S$. Consequently, we have $S \subset \text{der} S$. Using Equation $(1)$, this is equivalent to $\text{der} S = \text{clr} S$.

I got stock right here! Can you give me a hint?

Hosein Rahnama
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    Alternatively, to clear away the trees to see the forest: Let $U$ and $\overline{U}$ be an open set and its closure. Since $\overline{U}$ is closed (immediate), we only need to show that every point of $\overline{U}$ is a limit point of $\overline{U}.$ In fact, more is true, namely every point of $\overline{U}$ is a limit point of $U.$ To see this, note that every point of an open set is a limit point of that open set (easy) and every point of the closure of a set is a limit point of that set (easy or by definition, depending on how closure is defined). – Dave L. Renfro Mar 02 '20 at 11:43
  • @DaveL.Renfro: Thanks for the attention. I would be happy to have your clarification as an answer. :) – Hosein Rahnama Mar 02 '20 at 11:50
  • @DaveL.Renfro: Well, I thought a little and it appeared to me that we have open sets with no accumulation points. Consider the metric space $\mathbb{R}$ with the well-known discrete metric $d$. Then the singleton ${x}$ is open in $(\mathbb{R},d)$ but it does not have any accumulation points. Am I right? – Hosein Rahnama Mar 02 '20 at 12:10
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    I'm always a bit wary of terms like "limit point" and "accumulation point" and the like because their meanings can vary in the literature. If we're using the notion in which every open ball containing $x$ must contain a point in the set different from $x,$ then an open singleton set will not have any such points, and perhaps my de-cluttered version only works for metric spaces without isolated points. I don't have time now to sort all this out, but I do recommend Sections 10-13 in Point Sets by Eduard Čech (1969) as a nice reference. – Dave L. Renfro Mar 02 '20 at 13:15

2 Answers2

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You have shown that $\text{clr }S = \text{der }S$. It follows that $$\text{clr }S = \text{der }S \subset \text{der clr }S$$

where the last set inclusion is an instance of $A \subset B \implies \text{der }A \subset \text{der }B,$ which is true for any subsets $A,B$ of a given metric space

Brian Moehring
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Here is another way of showing this. In a general metric space, one can prove that

$$\text{clr}\,\text{der}S=\text{der}\,\text{clr}S=\text{der}S.$$

For example, take look at this post. To prove the theorem, one only needs to show

$$\text{clr}S=\text{der}\,\text{clr}S.$$

By the last paragraph in the question, we have $\text{clr}S=\text{der}S$ so the theorem follows immediately.

Hosein Rahnama
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